Question: Let $h(x)=x^3-9x^2+7x$ and let $c$ be the number that satisfies the Mean Value Theorem for $h$ on the interval $-3\leq x \leq6$. What is $c$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $-1$ (Choice B) B $0$ (Choice C) C $3$ (Choice D) D $4$
According to the Mean Value Theorem, there exists a number $c$ in the open interval $-3<x<6$ such that $h'(c)$ is equal to the average rate of change of $h$ over the interval: $h'(c)=\dfrac{h(6)-h(-3)}{(6)-(-3)}$ First, let's find that average rate of change: $\dfrac{h(6)-h(-3)}{(6)-(-3)}=\dfrac{-66-(-129)}{9}={7}$ Now, let's differentiate $h$ and find the $x$ -value for which $h'(x)={7}$. $h'(x)=3x^2-18x+7$ The solutions of $h'(x)=7$ are $x=0$ and $x=6$. Out of these, only $x=0$ is within the interval $-3<x<6$. In conclusion, $c=0$.